Помогите решить: 1) (sin3a cos^3a + cos3a sin^3a)/sin4a= 2) cos2a/(1-sin2a) - (1+tga)/(1-tga)=. Created by Пазенко. algebra-ru

(1-tga)=...

$\displaystyle \frac{(sin3a*cos^3a+cos3a*sin^3a)}{sin4a}=\frac{(3sina-4sin^3a)*cos^3a+(4cos^3a-3cosa)*sin^3a}{2sin2a*cos2a}=\\\\=\frac{3sina*cos^3a-4sin^3a*cos^3a+4cos^3a*sin^3a-3cosa*sin^3a}{4sina*cosa*cos2a}=\\\\=\frac{3sina*cosa(cos^2a-sin^2a)}{4sina*cosa*cos2a}=\frac{3}{4}$

$\displaystyle \frac{cos2a}{1-sin2a}-\frac{1+tga}{1-tga}=\frac{cos2a}{1-sin2a}-\frac{1+\frac{sina}{cosa}}{1-\frac{sina}{cosa}}=\frac{cos2a}{1-sin2a}-\frac{cosa+sina}{cosa-sina}=\\\\=\frac{cos^2a-sin^2a}{1-sin2a}-\frac{cosa+sina}{cosa-sina}=\\\\=(cosa+sina)*(\frac{cosa-sina}{1-sin2a}-\frac{1}{cosa-sina})=\\\\=(cosa+sina)*(\frac{cos^2a-2cosa*sina+sin^2a-1+sin2a}{(1-sin2a)(cosa-sina)})=\\\\=(cosa+sina)*(\frac{0}{(1-sin2a)(cosa-sina)})=0$

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